Rayner-Canham 5e Answers to Odd-Numbered Questions

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    Descriptive Inorganic Chemistry, Fifth Edition Answers to Odd-Numbered Questions 1

    2010 W. H. Freeman and Company, All Rights Reserved

    Chapter 1

    Exercises

    1.1 (a) Surface where electron probability = 0.

    (b) No two electrons in an atom can have the same 4

    quantum numbers.

    (c) Attraction into a magnetic field by an unpaired

    electron.

    1.3

    1.5 5p.

    1.7 Size of an orbital.

    1.9 With parallel spins there is zero probability that the

    electrons will occupy the same volume of space.

    1.11 (a) [Ne]3s1; (b) [Ar]4s23d8; (c) [Ar]4s13d10.

    1.13 (a) [Ar]; (b) [Ar]; (c) [Ar]3d9.

    1.15 1+ and 3+. Configuration of

    [Xe]6s24f145d106p1.

    1.17 1+. Configuration of [Kr]5s14d10.

    1.19

    1.21 E113: [Rn]7s25f1 46d1 07p 1 .E113 as +1 ion: [Rn]7s 25f1 46d1 0 .

    E113 as +3 ion: [Rn]5f1 46d1 0 .

    Beyond the Basics

    1.23 9, 5, 121.

    1.25 There are sevenforbitals. There are two separate

    ways of depicting them and designating them: the

    general set and the cubic set.

    1.27 Hydrogen heads the alkali metal group. Helium

    heads the alkaline earth metal group.

    Chapter 2

    Exercises

    2.1 (a) Lanthanum through lutetium.

    (b) Apparent radius of an atom in non-bonded contact

    with another.(c) Actual nuclear charge experienced by an electron.

    2.3 Argon did not fit into any of the then-known groups.Because the table was based on measured atomic

    mass, argon should have been placed between

    potassium and calcium.

    2.5 The long form correctly depicts the order ofelements; but the table becomes very elongated.

    2.7 The -iumending indicates a metal. The ending -on

    has been used for non-metals.

    2.9 With nuclei up to 26 protons, nuclear fusion is an

    exothermic process.

    2.11 (a) Lead (b) technetium (c) bromine.

    2.13 Sodium, because it has an odd number of protons.

    2.15 50.

    2.17 (a) Several nonmetals have metallic luster.

    (b) Diamond has the highest thermal conductivity ofall substances.

    (c) Graphite is a good electrical conductor in two

    dimensions.

    2.19 Potassium. As the effective atomic charge on the

    outermost electrons increases across a period.

    2.21 The effective nuclear charge on the 4pelectrons willbe increased.

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    4 Answers to Odd-Numbered Questions Descriptive Inorganic Chemistry, Fifth Edition

    2010 W. H. Freeman and Company, All Rights Reserved

    4.3 High electrical conductivity, high thermal

    conductivity, high reflectivity, and high boiling point.

    4.5 3sand 3pband overlap means that electrons in the

    full 3sband can spill over into the 3pband.

    4.7 For metallic behavior, the orbitals of the atoms must

    overlap.

    4.9 Cubic and hexagonal. Hexagonal.

    4.11 Simple cubic unit cell contains 4 atoms.

    4.13 Same size atoms, adopt the same structure, must have

    similar properties.

    Beyond the Basics

    4.15 The volume of the atom will be 4/3r3, while the

    volume of the cube will be (2r)3. The ratio of these

    gives 0.52.

    4.17 The length of the unit cell edge will be [4/(2)]r=

    2.83r.

    4.19 (a) 125 pm.

    (b) 7.24 gcm3.

    4.21 145pm.

    4.23 A suspension of gold nano-particles has a red color.

    Chapter 5

    Exercises5.1 (a) Distortion from a spherical shape.

    (b) The holes between anions in the crystal packing.(c) The diagram used to show the three bonding

    categories: metallic, covalent, and ionic.

    5.3 Hard and brittle crystals; high melting points;

    electrically conducting in liquid phase and in aqueous

    solution.

    5.5 (a) K+, because the radius will be determined by the

    inner orbitals.

    (b) Ca2+, because the ions are isoelectronic but

    calcium has one more proton.(c) Rb+, because again the ions are isoelectronic, with

    rubidium having two more protons than bromide.

    5.7 NaCl, because chloride is smaller than iodide; thecharge is more concentrated, and the ionic attraction

    will be stronger.

    5.9 Ag+, because it has the lowest charge density.

    5.11 UCl3(837C); UCl4(590C); UCl5(327C); UCl6

    (179). In this particular series, there does not appear

    to be a clear divide between high (ionic) and low

    (covalent) melting points.

    5.13 WF6(2C) and WO3(1472C). The fluoride is

    predominantly covalent while the oxide is ionic. Thefluoride is more polarizable than the oxide.

    5.15 Tin(II) chloride has a higher melting point because

    tin(II) has a fairly low charge density.

    5.17 No, ionic compounds do not dissolve in nonpolar

    solvents.

    5.19 Magnesium chloride, because the dipositive smaller

    magnesium ion has a significantly higher charge

    density.

    5.21 Lithium nitrate, because the lithium ion has a higher

    charge density than the sodium ion.

    5.23 The coordination number depends on the radius ratio

    5.25 The magnesium ion is smaller than the calcium ion.

    5.27

    5.29 (a) Metallic and a lesser contribution of ionic; (b)

    covalent and a lesser contribution of ionic.

    5.31 The choice would be metallic or covalent.

    Beyond the Basics

    5.33 Direct hybrid ionic-covalent bonding between pairs

    of ions in the gas.

    5.35 (a) Copper(II) chloride. The higher charge density

    copper(II) ion.

    (b) Lead(II) chloride. The very high charge density ofthe lead(IV) ion.

    5.37 1.15(r++r).

    5.39 164 pm

    5.41 251 pm

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    6 Answers to Odd-Numbered Questions Descriptive Inorganic Chemistry, Fifth Edition

    2010 W. H. Freeman and Company, All Rights Reserved

    For barium sulfate: 1044 kJmol1

    Yes.

    6.39

    For magnesium: H = +428 kJmol1. For lead: H =

    +898 kJmol1The only significant difference between the two ions

    is their hydration enthalpies.

    6.41 Using the Kapustinskii equation: U = 622 kJmol1;

    compared with 668 kJmol1experimentally and

    636 kJmol1from the Born-Land equation.

    6.43 H = +36 kJmol1

    6.45 proton affinity = 1141 kJmol1

    6.46 Step 1: G= +184 kJmol1

    Step 2: G= 47 kJmol1

    6.47 Ionic bond formation is best considered as acompetition for electrons.

    Chapter 7

    Exercises

    7.1 Polar protic: solvents with a dielectric constant

    between 50 and 100.Dipolar aprotic: solvents with dielectric constant

    between 20 and 50.

    Nonpolar: solvents with dielectric constant close tozero.

    7.3 (a) H+

    (aq) + OH

    (aq) H2O(l)(b) 2 HCO3

    (aq) + Co2+(aq) CoCO3(s) + H2O(l) +CO2(g)

    (c) OH(aq) + CH3COOH(aq) CH3COO(aq) +

    H2O(l)

    7.5 (a) Pairs of species that differ in formula by one

    ionizable hydrogen.(b) Solvent that undergoes its own acid-base reaction.

    (c) Ability of a substance to act as an acid or a base.

    7.7 (a) NH4+(aq) + H2O(l) NH3(aq) + H3O

    +(aq)

    (b) CN(aq) + H2O(l) HCN(aq) + OH(aq)

    (c) HSO4(aq) + H2O(l) SO4

    2(aq) + H3O+(aq)

    7.9 ClNH2(aq) + H2O(l) ClNH3+(aq) + OH(aq)

    7.11 H2SO4(l) + H2SO4(l) H3SO4+(H2SO4) +

    HSO4(H2SO4)

    7.13 (a) The ammonium ion, NH4+; (b) the amide ion,

    NH2.

    7.15 HF(H2SO4) + H2SO4(l) H2F+(H2SO4) +

    HSO4(H2SO4)

    HF is acting as a base and H2F+is the conjugate

    acid,

    7.17 HSeO4(base), H2SO4(conjugate acid); H2O

    (acid), OH(conjugate base).

    7.19 The HSe bond will be weaker than the HS bond.

    Thus hydrogen selenide will be the stronger acid.

    7.21 [Zn(OH2)6]2+

    (aq) + H2O(l)

    [Zn(OH2)5(OH)]+(aq) + H3O

    +(aq)

    7.23 The diprotic acid must be present in the least

    proportion.

    H2NNH2(aq) + H2O(l) H2NNH3+(aq) + OH(aq)

    H2NNH3+(aq) + H2O(l)

    +H3NNH3+(aq) +

    OH(aq)

    7.25 (a) Acidic, because aluminum is a small high-charge

    cation.

    (b) Neutral, because the sodium ion will stay

    unchanged.

    7.27 With a smaller pKb, Amust be the stronger base.

    Thus HB will be the stronger acid.

    7.29 H3PO4(aq) + HPO42(aq) 2 H2PO4

    (aq)

    7.31 (a) N2O5; (b) CrO3; (c) I2O7.

    7.33 (a) SiO2(acid), Na2O (base); (b) NOF (acid), ClF3(base); (c) Al2Cl6(acid), PF3(base).

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    Descriptive Inorganic Chemistry, Fifth Edition Answers to Odd-Numbered Questions 7

    2010 W. H. Freeman and Company, All Rights Reserved

    7.35 (a) No effect.

    (b) Increasing pH. Se2(aq) + H2O(l)

    HSe(aq) + OH(aq)

    (c) Decreasing pH.

    [Sc(OH2)6]3+(aq) + H2O(l)

    [Sc(OH2)5(OH)]

    2+

    (aq) + H3O

    +

    (aq)(d) Increasing pH. F(aq) + H2O(l) HF(aq) +

    OH(aq)

    7.37 (a) Weakly basic; (b) neutral; (c) moderately basic;

    (d) strongly basic.

    7.39 (a) Strongly basic; (b) very strongly basic.

    7.41 G = 28 kJmol1. Magnesium oxide will be a

    weaker base than calcium oxide (G = 59

    kJmol1).

    7.43 NO+is a Lewis acid. Clis a Lewis base.

    (NO)(AlCl4)(NOCl) + [(CH3)4N]Cl(NOCl)

    [(CH3)4N](AlCl4)(NOCl) + NOCl(l)

    7.45 (a) [NH4+] = 3 1017molL1

    (b) [NH4+] = 11033molL1

    7.47 (a) No. The reactants have the combinations

    borderline-borderline and hard-hard.

    (b) Yes. The products will be preferred where hard-

    hard and soft-soft combinations result.

    7.49 (a) Greater than 1.

    (b) Less than 1.

    7.51 (a) Thallium(I), in analysis group I.

    (b) Rubidium ion, in analysis group V.(c) Radium ion, analysis group IV.

    (d) Iron(III), in the analysis group III.

    7.53 (a) MgSO4; (b) CoS.

    Beyond the Basics

    7.55 [S2-

    ] = 1.1 x 10-22

    . Cadmium sulfide will precipitate.Iron(II) sulfide will

    not precipitate.

    7.57 Zinc is a borderline acid, so it can be found as ores ofboth hard and soft bases.

    7.59 H2CO3(aq) + MgSiO4(s) H2O(l) + SiO2(s) +MgCO3(s)

    The atmospheric concentration of carbon dioxide hasdecreased, in part due to the formation of magnesium

    carbonate minerals.

    7.61 Dimethylsulfoxide must be a softer base than water.

    7.63 In terms of the HSAB concept, the harder calcium

    ion is likely to form a stronger bond to the water

    molecules of hydration than the softer barium ion.

    7.65 Fe

    3+

    (AlO6)

    3

    .

    Chapter 8

    Exercises

    8.1 (a) A substance that will oxidize another.

    (b) A two-dimensional plot of free energy against

    temperature for series of reactions that involve

    elements and their oxides, sulfides, or chlorides.

    8.3 (a) +3 (b) +5 (c) 3 (d) 3 (e) +5

    8.5 (a) 2; (b) +2; (c) 1; (d) +6; (e) 2.

    8.7 1, +1, +3, +5, +7.

    8.9 (a) +1; (b) +2; (c) +3; (d) +4; (e) +5. An increase by

    units of +1 from Group 13 to Group 17.

    8.11 (a) Nickel from +2 to 0, carbon from 0 to +2.

    (b) Manganese from +7 to +2, sulfur from +4 to +6.

    8.13 NH4+(aq) + 3 H2O(l) NO3

    (aq) + 10 H+(aq) + 8 e

    8.15 N2H4(aq) + 4 OH(aq) N2(g) + 4 H2O(l) + 4 e

    8.17 (a) 5 HBr(aq) + HBrO3(aq) 3 Br2(aq) + 3 H2O(l)

    (b) 2 HNO3(aq) + Cu(s) + 2 H+(aq)

    2 NO2(g) + Cu2+(aq) + 2 H2O(l)

    8.19 (a) 12 V(s) + 10 ClO3(aq) + 18 OH(aq)

    6 HV2O73(aq) + 10 Cl(aq) + 6 H2O(l)

    (b) 2 S2O42(aq) + 3 O2(g) + 4 OH

    (aq)

    4 SO42(aq) + 2 H2O(l)

    8.21 (a) Spontaneous.(b) Non-spontaneous.

    8.23 One example: Zn Zn2+, E = +0.762 V

    8.25 (a) Au3+(aq) + 3 eAu(s) would be the strongeroxidizing agent.

    (b) Al(s) Al3+(aq) + 3 ewould be the stronger

    reducing agent.

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    8 Answers to Odd-Numbered Questions Descriptive Inorganic Chemistry, Fifth Edition

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    8.27 E = 0.267 V

    8.29 E = +0.805 V

    8.31 (a) Br2

    (b) E = +0.52 V

    (c) The Nernst expression does not have a pH-dependent term.

    8.33 The most thermodynamically stable oxidation state is

    +3.

    8.35 At pH 0.00 because the reduction potential is lower.

    Beyond the Basics8.37 The oxidation to carbon monoxide involves an

    increase of entropy; thus the TS term will become

    increasingly negative with increase in temperature.

    The negative slope for this line will ultimately cross

    the carbon dioxide line.T = 766 K = 493C

    8.39 E will increase to the point where insoluble, brown

    manganese(III) oxide will be formed, thusdiscoloring the toilet bowl.

    Chapter 9

    Exercises

    9.1 (a) A pair of elements in a compound whose sum of

    valence electrons adds up to eight.

    (b) The relationship between an element and theelement to its lower right in the periodic table.

    9.3 The general formula: M+M3+(SO42)212H2O, whereM+is potassium or ammonium and M3+is aluminum,chromium(III), or iron(III).

    9.5 KF, CaF2, GaF3, GeF4, AsF5, SeF6, BrF5, KrF2.

    The bonding in the potassium and calcium fluorides

    is ionic, while that for the germanium, arsenic,selenium, bromine, and krypton compounds is

    covalent.

    9.7 (a) Hydrogen gas, H2; (b) calcium metal.

    9.9 As the group is descended, the cation radii increase,

    the ionic bond will weaken and the melting point will

    be lower.

    9.11 Scandium hydroxide, Sc(OH)3.

    9.13 SO3(s) + H2O(l) H2SO4(aq)CrO3(s) + H2O(l) H2CrO4(aq)

    9.15 (a) Al2O3, Sc2O3.

    (b) P2O5, V2O5.

    9.17 Tin.

    9.19 N2, O2, F2; thus they have stronger dispersion

    (London) forces.

    9.21 Forming the Eu2+ion would retain the half-filled d

    orbital set.

    9.23 (a) Indium(III) and bismuth(III); (b) cadmium(II)

    and lead(II).

    9.25 Thallium(I) bromide.

    9.27 (a) CO; (b) (CC)2.

    9.29 Yttrium.

    Beyond the Basics

    9.31 Because the synthetic route involves a negative free

    energy change.

    9.33 Add excess hydroxide ion.

    9.35 (a) 12, (b) 7.

    9.37 Li (+1); Be (+2); B (+3); C (+4); N (+3); O (+2). For

    Period 2, oxidation numbers reach a maximum at

    carbon, then decrease.Na (+1); Mg (+2); Al (+3); Si (+4); P (+5); S (+6); Cl

    (+5). For Period 3, the oxidation number matches the

    number of valence electrons except for chlorine.

    9.39 Fe2O3(+3); RuO4(+8); OsO4(+8). For ruthenium

    and osmium, the oxidation number is the same as the

    Group number.

    9.41

    Group 15 Group 16 Group 17

    Group 15 CN22 OCN

    FCN

    Group 16 OCN CO2 FCO+

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    Descriptive Inorganic Chemistry, Fifth Edition Answers to Odd-Numbered Questions 9

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    Chapter 10

    Exercises

    10.1 (a) A hydrogen atom bridging atoms in a covalent

    bond in which the hydrogen is less electronegative.(b) A hydrogen atom bridging atoms in a covalent

    bond in which the hydrogen is more electronegative.

    10.3 The ice cube consists of heavy water, deuterium

    oxide.

    10.5 The difference in absorption frequency is very small,

    about 106of the signal itself.

    10.7 Hydrogen rarely forms a negative ion.

    10.9 Enthalpy driven. The chemical equation is: N2(g) +

    3 H2(g) 2 NH3(g)

    There is a decrease in the number of gas molecules,hence a decrease in entropy.

    10.11 (a) 2 KHCO3(s) K2CO3(s) + H2O(g) + CO2(g)

    (b) HCCH(g) + 2 H2(g) H3CCH3(g)

    (c) PbO2(s) + 2 H2(g) Pb(s) + 2 H2O(g)

    (d) CaH2(s) + H2O(l) Ca(OH)2(aq) + H2(g)

    10.13 The much lesser enthalpy of formation of ammoniacompared to water can be explained in terms of the

    much greater bond energy of dinitrogen (945

    kJmol1) compared with that of dioxygen (498

    kJmol1).

    10.15 There are three categories of covalent hydrides: those

    in which the hydrogen is nearly neutral; those in

    which it is quite positive, and those in which it is

    negative. Most covalent hydrides belong in the first

    category.

    10.17 KH; CaH2, GaH3, GeH4, AsH3, H2Se, HBr. The trend

    is to increase by one H until germanium, then astepwise decrease by one H to hydrogen bromide.

    10.19 (a) Gas. It is a covalent hydride.(b) Solid. This is an ionic hydride.

    10.21 The closeness of the electronegativities of hydrogen

    and carbon, and the ability to hydrogen bond.

    Beyond the Basics

    10.23 (a) Yes, liquid; (b) no, gas; (c) yes, liquid; (d) no

    gas.

    10.25. Looking at a generic Born-Haber cycle, where X = H

    or Cl, we see that there are two features that differ.

    10.27 Hydrogen and carbon monoxide.H2O(l) + C(s) H2(g) + CO(g)

    The combustion reaction would therefore be:

    H2(g) + CO(g) + O2(g) H2O(g) + CO2(g)

    H = 525 kJ

    Per mole, this is = 262 kJmol1, compared with

    242 kJmol1for the combustion of pure

    dihydrogen.

    Chapter 11

    Exercises

    11.1 (a) 2 Na(s) + 2 H2O(l) 2 NaOH(aq) + H2(g)

    (b) Rb(s) + O2(g) RbO2(s)

    (c) 2 KOH(s) + CO2(g) K2CO3(s) + H2O(l)

    (d) 2 NaNO3(s) 2 NaNO2(s) + O2(g)

    11.3 They resemble typical metals in that they are shiny

    and silvery and good conductors of heat andelectricity. The alkali metals differ from typical

    metals in that they are soft, extremely chemically

    reactive, have low melting points and very lowdensities.

    11.5 All common chemical compounds are water soluble.

    They always form ions of +1 oxidation state.

    Their compounds are almost always ionic.

    11.7 The most likely argument is that the hydroxide ion

    can hydrogen bond with the surrounding watermolecules.

    11.9 Because the equilibrium of the synthesis reaction:

    Na(l) + KCl(l) K(l) + NaCl(l) lies to the left.

    Hydrogen Chlorine

    Bond energy 432 kJmol1 240 kJmol1

    Electron affinity 79 kJmol1 349 kJmol1

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    10 Answers to Odd-Numbered Questions Descriptive Inorganic Chemistry, Fifth Edition

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    11.11 (a) Sodium hydroxide; (b) anhydrous sodium

    carbonate; (c) sodium carbonate decahydrate.

    11.13 (a) Loss of water by a hydrated salt in a low-humidity

    environment.(b) Chemical similarities of one element and the

    element to its lower right in the periodic table.

    11.15 CO2(g) + NH3(aq) + H2O(l) NH4+(aq) +

    HCO3(aq)

    HCO3(aq) + Na+(aq) NaHCO3(s)

    2 NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)

    CaCO3(s) CaO(s) + CO2(g)

    CaO(s) + H2O(l) Ca(OH)2(s)

    2 NH4+(aq) + 2 Cl(aq) + Ca(OH)2(s)

    2 NH3(aq) + CaCl2(aq) + 2 H2O(l)

    The problems: disposal of waste calcium chloride,and the high energy requirements.

    11.17 The ammonium ion is monopositive; its salts are all

    soluble; its size is about the middle of the alkali metalion range; all its common salts are colorless.

    11.19 Potassium dioxide(1) has a lower molar mass, and ischeaper.

    11.21 The ammonium ion is large.

    11.23 Lithium:

    6 Li(s) + N2(g) 2 Li3N(s)2 Li(s) + Cl2(g) 2 LiCl(s)

    Li(s) + C4H9Cl(solv) LiC4H9(solv) + LiCl(s)

    4 Li(s) + O2(g) 2 Li2O(s)

    2 Li(s) + H2O(l) 2 LiOH(aq) + H2(g)

    Li2O(s) + H2O(l) 2 LiOH(aq)

    2 LiOH(aq) + CO2(g) Li2CO3(aq) + H2O(l)

    Li2O(s) + CO2(g) Li2CO3(s)

    Sodium:

    2 Na(s) + Cl2(g) 2 NaCl(s)

    2 Na(s) + H2O(l) 2 NaOH(aq) + H2(g)

    2 Na(s) + O2(g) Na2O2(s)

    Na2O2(g) + H2O(l) 2 NaOH(aq) + H2O2(aq)

    2 NaOH(aq) + CO2(g) Na2CO3(aq) + H2O(l)Na2CO3(aq) + CO2(g) + H2O(l) 2 NaHCO3(aq)

    2 Na(s) + 2 NH3(l) 2 NaNH2(NH3) + H2(g)

    Na2O2(s) + CO2(g) Na2CO3(s) + O2(g)Potassium

    Na(l) + KCl(l) K(g) + NaCl(l)

    2 K(s) + Cl2(g) 2 KCl(s)

    K(s) + O2(g) KO2(s)

    2 KO2(s) + 2 H2O(l) 2 KOH(aq) + H2O2(aq) +

    O2(g)

    2 K(s) + 2 H2O(l) 2 KOH(aq) + H2(g)

    2 KOH(aq) + CO2(g) K2CO3(aq) + H2O(l)

    K2CO3(aq) + CO2(g) + H2O(l) 2 KHCO3(aq)

    2 KO2(s) + CO2(g) K2CO3(s) + 2 O2(g)

    3 K+(aq) + [Co(NO2)6]3(aq) K3[Co(NO2)6](s)

    Beyond the Basics11.25 Current = 6.94 x 104A

    11.27 In the series LiF to CsF, there is an increasing

    mismatch in ion sizes. For the series LiI to CsI, thereis a decreasing mismatch in ion sizes.

    11.29 NaBF4. The hydration energy will more probably

    exceed the (lower) lattice energy, making thecompound more soluble.

    11.31 Either: that there is appreciable covalent bonding inthe lithium hydride, or that the lithium ion is so small

    that the lattice consists of touching hydride ions with

    lithium ions rattling around in the lattice holes..

    11.33 LiF and KI.

    11.35 Calcium-40 is a doubly magic nucleus with filled

    shells of protons and neutrons.

    Chapter 12

    Exercises

    12.1 (a) 2 Ca(s) + O2(g) 2 CaO(s)

    (b) CaCO3(s) CaO(s) + CO2(g)

    (c) Ca(HCO3)2(aq) CaCO3(s) + H2O(l) +CO2(g)

    (d) CaO(s) + 3 C(s) CaC2(s) + CO(g)

    12.3 (a) Barium; (b) barium.

    12.5 The higher charge density magnesium ion will cause

    the water molecules surrounding it during the

    hydration step to become much more ordered than

    with the lower charge density sodium ion.

    12.7 They form 2+ ions exclusively and their salts tend to

    be highly hydrated.

    12.9 Steric hindrance.

    12.11 Rainwater, an aqueous solution of carbon dioxide,

    percolates into limestone deposits, reacting with thecalcium carbonate to give a solution of calcium

    hydrogen carbonate.

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    12.13 Ca(OH)2(aq) + Mg2+(aq) Mg(OH)2(s) +

    Ca2+(aq)

    Mg(OH)2(s) + 2 HCl(aq) MgCl2(aq) + 2 H2O(l)

    Mg2+(MgCl2) + 2 eMg(l)

    2 Cl-(MgCl2) Cl2(g) + 2 e

    12.15 (a) Ca(OH)2(hydrated lime) or CaO (quicklime); (b)

    Mg(OH)2;

    (c) MgSO47 H2O.

    12.17 Lead is used because it has the highest atomic

    number of the common, non-radioactive elements.

    12.19 Both form tough oxide coatings over their surface;

    they are amphoteric, forming beryllate and aluminate

    anions; they form carbides containing the C4ion.

    12.21 Magnesium ion is a key component of chlorophyll.

    12.23 (a) Mg(s) + HCl(aq) MgCl2(aq) + H2(g)then evaporate to crystallize MgCl26

    H2O(s).

    (b) Mg(s) + Cl2(g) MgCl2(s)

    Beyond the Basics

    12.25 H = +83 kJmol1

    S = 0.220 kJmol1K1T = 377 K

    12.27 The formula is actually [Mg(OH2)6]2+

    [SO4H2O]2

    .

    12.29 BeH+. This ion would possess a single bond.

    12.31 Ca3N2(s) + 4 NH3(l) 3 Ca(NH2)2(NH3)

    12.33 G = 92 kJmol1. Less favorable, for at a higher

    temperature, the low- melting magnesium will be aliquid. The reason for synthesizing at a higher

    temperature is the greatly increased rate of reaction.

    12.35 The species is probably Na2BeCl4,.

    Chapter 13

    Exercises

    13.1 (a) 3 K(l) + AlCl3(s) Al(s) + 3 KCl(s)

    (b) B2O3(s) + 2 NH3(g) 2 BN(s) + 3 H2O(g)

    (c) 2 Al(s) + 2 OH(aq) + 6 H2O(l) 2

    [Al(OH)4](aq) + 3 H2(g)

    (d) 2 B4H10(g) + 11 O2(g) 4 B2O3(s) + 10

    H2O(g)

    13.3 The bridging oxygen atoms have an oxidation

    number of 1:

    13.5 An arachno-cluster.

    13.7 -1,042kJ. The major factors are the weak fluorine-

    fluorine bond, and the exceedingly strong boron-

    fluorine bond.

    13.9 Al3+is surrounded by the partially negative oxygen

    atoms of the six water molecules.

    13.11 The hydrated aluminum ion acts as a Bronsted-Lowry

    acid.

    13.13 The potential environmental hazards are red mud;hydrogen fluoride gas; the carbon oxides; and

    fluorocarbon compounds produced.

    13.15 Aluminum fluoride is a typical ionic compound.

    Both aluminum bromide and aluminum iodide are

    covalently bonded dimers. Aluminum chloride is a

    borderline case.

    13.17 A spinel has the formula AB2X4, where A is a

    dipositive metal ion, B is a tripositive metal ion, and

    X is a dinegative ion. In the reverse spinel, the A

    cations occupy octahedral sites while half of the Bcations occupy the tetrahedral sites.

    13.19 Gallium(III) fluoride must consist of an ionic lattice

    of gallium(3+) and chloride(1) ions.

    13.21 In acid conditions, the soluble Al(OH2)63+is

    produced. The aluminum ion is very toxic to fish.

    Beyond the Basics

    13.23 The metallic radius is a measure of the atomic size.

    The covalent radius will be smaller because there is

    orbital overlap. The ionic radius is by far thesmallest because all the valence electrons have been

    lost.

    13.25 Cl3Al[O(C2H5)2].

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    13.27 The beryllium ion will resemble the aluminum ion.

    [Be(OH2)4]2+(aq) + H2O(l) [Be(OH2)3(OH)]

    +(aq)

    + H3O+(aq)

    13.29 Let number of ions of magnesium =x, then:x= +3.

    13.31 3 GaCl(s) GaCl3(s) + 2 Ga(s)There are equal moles (in the same phase) on eachside of the equation.

    13.33 4 AlCl3(s) + CH3CN(l) [Al(CH3CN)6]3+(CH3CN)

    + 3 [AlCl4](CH3CN)

    13.35 Ga(OH2)63+(aq) GaO(OH)(s) + H2O(l) + 3

    H3O+(aq)

    Addition of acid will shift the equilibrium to the left.

    13.37 Aluminum, lacking any inner delectrons, behavesmore like a Group 3 element than a Group 13

    element.

    13.39

    13.41 208 kJmol1. B.O. = 1.

    13.43 Hf(B2O3) = 1271 kJmol1

    13.45 Using the atomic radius of zirconium would give a

    ratio of sizes of close to unity: not an NaCl packing

    pattern. The structure must be [Zr4+][B12

    4].

    Chapter 14

    Exercises

    14.1 (a) Li2C2(s) + 2 H2O(l) 2 LiOH(aq) + C2H2(g)

    (b) SiO2(s) + 2 C(s) Si(l) + 2 CO(g)

    (c) CuO(s) + CO(g) Cu(s) + CO2(g)

    (d) Ca(OH)2(aq) + CO2(g) CaCO3(s) + H2O(l)

    CaCO3(s) + CO2(g) + H2O(l) Ca(HCO3)2(aq)

    (e) CH4(g) + 4 S(l) CS2(g) + 2 H2S(g)

    (f) SiO2(s) + 2 Na2CO3(l) Na4SiO4(s) + 2

    CO2(g)

    (g) PbO2(s) + 4 HCl(aq) PbCl4(aq) + 2 H2O(l)

    PbCl4(aq) PbCl2(s) + Cl2(g)

    14.3 (a) An element forming chains of its atoms.(b) Low density silicates with numerous cavities in

    the structure.

    (c) Non-metallic inorganic compounds.

    (d) Chains of alternating silicon and oxygen atomswith organic side groups.

    14.5 Diamond is a very hard, transparent, colorless solid

    that is a good conductor of heat but a non-conductorof electricity. Graphite is a soft, slippery, black solid

    that is a poor conductor of heat but a good conductor

    of electricity. C60is black and a nonconductor of

    heat and electricity.

    14.7 Diamond and graphite both have network covalent

    bonded structures. The solvation process cannot

    provide the energy necessary to break nonpolar

    covalent bonds. The fullerenes consist of discretemolecules, such as C60. These individual nonpolarunits can become solvated by nonpolar or low-

    polarity solvent molecules and hence dissolve.

    14.9 The three classes are ionic, covalent, and metallic.

    14.11 SiO2(s) + 3 C(s) SiC(s) + 2 CO(g), entropy

    driven.

    H = +624 kJmol1

    S = +0.354 kJmol1K1

    G = 181 kJmol1

    14.13 It is the lower bond energy of the C=S bond

    compared to the C=O bond that makes such a large

    difference.

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    14.15 sphybrid orbitals are formed.

    14.17 Silicon in silane molecule has empty 3dorbitals that

    can be involved in the oxidation process.

    14.19 The synthesis of HFC-134a requires a complex,

    expensive multistep procedure.

    14.21 It absorbs wavelengths in the infrared region that are

    currently transparent.

    14.23

    14.25 Trigonal planar.

    14.27 There are three Fe2+ions and two Fe3+ions per

    formula.

    14.29 Zeolites are used as ion exchangers; as adsorptionagents; for gas separation; and as specialized

    catalysts.

    14.31 Any polymer molecules that leak in breast implants

    cannot be broken down by normal bodily processes.

    14.33

    14.35 PbO(s) + H2O(l) PbO2(s) + 2 H+(aq) + 2 e

    PbO(s) + 2 H+

    (aq) + 2 e

    Pb(s) + H2O(l)

    14.37 CNand CO.

    14.39 The lack of the range of synthetic pathways.

    14.41 Carbon:

    4 CO(g) + Ni(s) Ni(CO)4(g)

    CO(g) + Cl2(g) COCl2(g)

    CO(g) + S(s) COS(g)

    HCOOH(l) CO(g) + H2O(l) 42SOH

    CO2(g) + 2 Ca(s) C(s) + 2 CaO(s)

    2 CO(g) + O2(g) 2 CO

    2(g)

    CO(g) + 2 H2(g) CH3OH(l) catalyst

    2 C(s) + O2(g) 2 CO(g)

    C(s) + O2(g) CO2(g)

    Na2C2(s) + 2 H2O(l) 2 NaOH(aq) + C2H2(g)

    2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l)

    Al4C3(s) + H2O(l) 3 CH4(g) + 4 Al(OH)3(s)

    CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l)

    CH4(g) + 4 S(l) CS2(g) + 2 H2S(g)

    CS2(g) + 3 Cl2(g) CCl4(g) + S2Cl2(l)

    CS2(g) + S2Cl2(l) CCl4(g) + 6 S(s)

    CH4(g) + NH3(g) HCN(g) + 3 H2(g)

    HCN(aq) + H2O(l) H3O+(aq) + CN

    (aq)

    Ca(OH)2(aq) + CO2(g) CaCO3(s) + H2O(l)

    CaCO3(s) + 2 HCl(aq) CaCl2(aq) + H2O(l) +

    CO2(g)

    CaCO3(s) + H2O(l) + CO2(g) Ca(HCO3)2(aq)Silicon:

    Si(s) + HCl(g) SiHCl3(g) + H2(g)

    2 CH3Cl(g) + Si(s) (CH3)2SiCl2(l)

    SiO2(s) + 2 C(s) Si(s) + 2 CO(g)

    SiO2(s) + 6 HF(aq) SiF62(aq) + 2 H+(aq) + 2

    H2O(l)

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    SiO2(s) + 2 NaOH(l) Na2SiO3(s) + H2O(g)

    SiO2(s) + 3 C(s) SiC(s) + 2 CO(g)

    SiO2(s) + 2 Na2CO3(l) Na4SiO4(s) + 2 CO2(g)

    2 SiO44(aq) + 2 H+(aq) Si2O7

    6(aq) + H2O(l)

    Beyond the Basics

    14.43 Its a calcium ion mimic.

    14.45 Sodium and calcium ions can leach out.

    14.47 (a) A six-membered ring structure, Si3O3, withalternating silicon and oxygen atoms.

    (b) P3O93

    (c) S3O9.

    14.49 Tin(II) chloride is the Lewis acid, while the chloride

    ion, the Lewis base.

    14.51 Mg2SiO4(s) + 2 H2CO3(aq) 2 MgCO3(s) +

    SiO2(s) + 2 H2O(l)

    14.53 A = CH4; B = S; C = CS2; D = H2S; E = Cl2; F = CCl4

    CH4(g) + 4 S(s) CS2(g) + 2 H2S(g)

    CS2(g) + 2 Cl2(g) CCl4(g) + 2 S(s)

    CH4(g) + 4 Cl2(g) CCl4(g) + 4 HCl(g)

    14.55 Y is Sn(C2H5)4, Z is SnCl(C2H5)3

    3 Sn(C2H5)4(l) + SnCl4(l) 4 SnCl(C2H5)3(l)

    14.57 G = +48 kJmol1, a positive value indicates

    decomposition will be favored.

    14.59 Energy released = 394 kJmol1

    Chapter 15

    Exercises

    15.1 (a) AsCl3(l) + 3 H2O(l) H3AsO3(aq) + 3 HCl(g)

    (b) 3 Mg(s) + N2(g) Mg3N2(s)

    (c) NH3(g) + 3 Cl2(g) NCl3(l) + 3 HCl(g)

    (d) CH4(g) + H2O(g) CO(g) + 3 H2(g)

    (e) N2H4(l) + O2(g) N2(g) + 2 H2O(g)

    (f) NH4NO3(aq) N2O(g) + 2 H2O(l)

    (g) 2 NaOH(aq) + N2O3(aq) 2 NaNO2(aq) + H2O(l)(h) 2 NaNO3(s) 2 NaNO2(s) + O2(g)

    (i) P4O10(g) + C(s) P4(g) + 10 CO(g)

    15.3 Arsenic has both metallic and nonmetallic allotropes.

    15.5 Difference in boiling points; different acid-base

    properties; difference in their combustions.

    15.7 (a) Nitrogen has a very strong nitrogen-nitrogen triplebond.

    (b) Kinetic factors can lead to other products.

    15.9 Air. Cool the mixture and have the argon condenseout.

    15.11 A solution of the ion is acidic, not neutral, and its

    compounds are all very thermally unstable.

    15.13

    15.15 Volume of gas = 2.8 L

    15.17 Hydrogen bonding in ammonia molecules.

    15.19 The shapes are:

    15.21 High pressure favors the reaction direction that will

    result in the lesser moles of gas.

    15.23 White phosphorus is a very reactive, white, waxy

    substance that consists of P4, while red phosphorus isa red powdery solid that consists of long polymer

    chains.

    15.25 Ammonia must be the stronger base.

    15.27 NO bond order is 2:

    Net energy change = 53 kJ .

    15.29

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    15.31 Steric hindrance by bromine.

    15.33 In the azide ion, the double-double nitrogen-nitrogen

    bond is strongly preferred.

    15.35 NH2OH(aq) + BrO3(aq) NO3

    (aq) + Br(aq) +

    H3O+(aq)

    15.37 NOF(g) + SbF5(l) NO+(SbF5) + SbF6

    (SbF5)

    15.39 H2S2O7and H6Si2O7.

    15.41 (a) Rapid algae growth leading to a depletion ofdissolved dioxygen.

    (b) Mutually beneficial relationship between two

    organisms.

    (c) Use of a chemical compound to combat disease.(d) Calcium hydroxide phosphate that is the bone

    material.

    Beyond the Basics

    15.43 PH4+and Cl, then BCl4

    .

    PH3(g) + HCl(l) PH4+(HCl) + Cl(HCl)

    Cl(HCl) + BCl3(HCl) BCl4(HCl)

    15.45 Trigonal planar; 120; bond order would be 1.33 inthe first case and 1.17 in the other.

    15.47 The most obvious structure would be that in which

    the four terminal oxygen atoms in P4O10are replacedby sulfur atoms.

    15.49 Bonding between sodium and azide ions is likely to

    be predominantly ionic whereas that in the heavymetal azides will be more covalent.

    15.51 (a) K = 6 102

    (b) K = 7103

    (c) Equilibrium is attained much more rapidly.

    15.53 H = 57 kJmol1

    15.55 Mass Na2HPO4= 4.0 g, mass NaH2PO4= 8.6 g

    15.57 Assuming that the PCl bond has about the same

    energy in PCl5and PCl3, the dissociation energy is =

    412 kJmol1,

    For the decomposition of PF5, the energy change will

    be = 825 kJmol1.

    This much higher value results from fluorine bonds

    to other elements being stronger than those of

    chlorine to the same element..

    15.59 [NF4]+F

    15.61 [A] Red phosphorus; [B] white phosphorus; [C]tetraphosphorus decaoxide;

    [D] phosphoric acid; [E] phosphorus trichloride; [F]

    phosphorus pentachloride; [G]phosphorous/phosphonic acid.

    4 P(s) P4(s)

    P4(s) + 5 O2(g) P4O10(s)

    P4O10(s) + 6 H2O(l) 4 H3PO4(aq)

    P4(s) + 6 Cl2(g) 4 PCl3(l)

    PCl3(l) + Cl2(g) PCl5(s)

    PCl5(s) + 4 H2O(l) H3PO4(aq) + 5 HCl(g)

    PCl3(l) + 3 H2O(l) H3PO3(aq) + 3 HCl(g)

    15.63 Li3N(s) + 3 H2O(l) 3 LiOH(aq) + NH3(g)This would be uneconomical.

    15.65 HONH2(or NH2OH, hydroxylamine); H2NNO2;(NH2)2CO (urea).

    15.67 2 NCl3(g) N2(g) + 3 Cl2(g)The reaction is highly exothermic due primarily to

    the strength of the nitrogen-nitrogen triple bond.

    15.69 Only two hydrogen atoms are replaced because the

    structure contains only two hydroxyl groups.

    15.71 NO2+

    and CNO

    15.73 A very large low-charge anion might stabilize the

    pentanitrogen cation.

    15.75 (a) Silver(I) or lead(II) or mercury(I).

    (b) N3(aq) + H2O(l) HN3(aq) + OH

    (aq)

    (c) The azide ion will decompose on heating.

    15.77 3 (NH4)[N(NO2)2](s) + 4 Al(s) 2 Al2O3(s) + 6H2O(g) + 6 N2(g)

    Reasons for its exothermicity: (a) the formation of

    dinitrogen; (b) the formation of water; (c) the

    formation of aluminum oxide.

    It would be a good propellant because of the largevolume of gas produced per mole of ADN.

    15.79 Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g)

    9 H2(g) + 2 AsO42(aq) + 4 H+(aq) 2 AsH3(g) + 8

    H2O(l)

    2 AsH3(g) 2 As(s) + 3 H2(g)

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    Chapter 16

    Exercises

    16.1 (a) 2 Fe(s) + 3 O2(g) 2 Fe2O3(s)

    (b) BaS(s) + 4 O3(s) BaSO4(s) + 4 O2(g)

    (c) BaO2(s) + 2 H2O(l) Ba(OH)2(aq) + H2O2(aq)

    (d) 2 KOH(aq) + CO2(g) K2CO3(aq) + H2O(l)

    K2CO3(aq) + CO2(g) + H2O(l) 2 KHCO3(aq)

    (e) Na2S(aq) + H2SO4(aq) Na2SO4(aq) + H2S(g)

    (f) Na2SO3(aq) + H2SO4(aq) Na2SO4(aq) +

    SO2(g) + H2O(l)

    (g) 8 Na2SO3(aq) + S8(s) 8 Na2S2O3(aq)

    16.3 Its electrical resistivity is low enough to be

    considered metallic.

    16.5 (a) Finely divided metals that are spontaneously

    flammable in air.(b) Different crystal forms of an element.

    (c) Unusual type of equilibria found with hemoglobin

    in which addition of one oxygen molecule increasesthe ease of addition of subsequent oxygen molecules.

    16.7 Photosynthesis has resulted in the conversion to

    dioxygen of most of the carbon dioxide.

    16.9 Bond order, about 1.

    16.11 Larger. Because of steric crowding.

    16.13 The oxidation number of +1 for oxygen is a result ofeach atom being sandwiched between a more

    electronegative fluorine atom.

    16.15 Among the Group 16 elements, it is only sulfur that

    readily catenates.

    16.17 The structures are:

    16.19 The structure is probably based on the S8ring.

    16.21

    16.23 Ca(OH)2(s) + CO2(g) CaCO3(s) + H2O(l)

    16.25 At higher temperatures, S8rings break into S2

    molecules analogous to O2.

    16.27 The closeness of the bond angle in H2Te to 90

    suggests that the central tellurium atom is using pureporbitals in its bonding.

    16.29 Sulfuric acid can act as an acid; as a dehydratingagent, as an oxidizing agent, as a sulfonating agent,and as a base with stronger acids.

    16.31 Sulfur trioxide.

    16.33 The formal charge representations are:

    16.35 (a) H2S(g) + Pb(CH3COO)2(aq) PbS(s) + 2

    CH3COOH(aq)

    (b) Ba2+(aq) + SO42(aq) BaSO4(s)

    16.37 There is a very high activation energy barrier to the

    reaction SO2SO3.

    16.39 The large tetramethylammonium cation will stabilizethe large, low-charge ozonide ion.

    16.41 The NS2+ion is isoelectronic and isostructural with

    carbon disulfide, CS2.

    16.43 We require only small quantities of selenium for ahealthy existence.

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    Beyond the Basics

    16.45 The value of 668 kJ is much less than the 1209 kJ

    for sulfur hexafluoride. This difference is accountedfor by the chlorine-chlorine bond being stronger.

    16.47 The ammonium salt will be less basic than the

    calcium salt because the ammonium ion is theconjugate base of a weak acid.

    16.49 Concentration in ppb = 2 105ppb

    16.51 (a) Length of side = 400 pm

    (b) Thus length of side = 339 pm.

    16.53

    16.55 mass = 94 tonne

    16.57 Apparent oxidation number S: [S] = +8, an

    impossible value because the oxidation number ofsulfur cannot exceed 6.

    16.59 SO32(aq) + S2O8

    2(aq) + H2O(l) 3 SO42(aq) + 2

    H+(aq)

    16.61 E = 1.48 V

    16.63 [A] Sulfur dioxide; [B] potassium hydroxide; [C]potassium sulfite; [D] sulfur; [E] thiosulfate ion; [F]tetrathionate ion; [G] thiosulfuric acid.

    SO2(g) + 2 KOH(aq) K2SO3(aq)

    K+(aq) + [B(C6H5)4](aq) K[B(C6H5)4](s)

    K2SO3(aq) + S(s) K2S2O3(aq)

    2 S2O32(aq) + I2(aq) S4O6

    2(aq) + 2 I(aq)

    S2O32(aq) + 2 H+(aq) H2S2O3(aq)

    H2S2O3(aq) H2O(l) + S(s) + SO2(g)

    16.65 The triple-bond structure is more likely.

    16.67 Rubidium or cesium. A large low-charge cation is

    necessary.

    16.69

    16.71. The species would be isoelectronic and isostructuralwith the carbonate ion and the nitrate ion.

    Chapter 17

    Exercises

    17.1 (a) UO2(s) + 4 HF(g) UF4(s) + 2 H2O(l)

    (b) CaF2(s) + H2SO4(l) 2 HF(g) + CaSO4(s)

    (c) SCl4(l) + 2 H2O(l) SO2(g) + 4 HCl(g)

    (d) 3 Cl2(aq) + 6 NaOH(aq) NaClO3(aq) + 5NaCl(s) + 3 H2O(l)

    (e) I2(s) + 5 F2(g) 2 IF5(s)

    (f) BrCl3(l) + 2 H2O(l) 3 HCl(aq) + HBrO2(aq)

    17.3 Fluorine has a very weak fluorine-fluorine bond; itscompounds with metals are often ionic when those of

    the comparable chlorides are covalent; it forms the

    strongest hydrogen bonds known; it tends to stabilize

    high oxidation states; the solubility of its metal

    compounds is often quite different than those of theother halides.

    17.5 The reaction with nonmetals is strongly enthalpy-

    driven.

    17.7 I2(s) + 7 F2(g) 2 IF7(s) There is a decrease of

    seven moles of gas in this reaction..

    17.9 Because hydrogen ion does not appear in the half-equation, the reduction potential will not be pH

    sensitive.

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    17.11 The HF bond is particularly strong..

    17.13 Mass of calcium sulfate = 4.1 1012g = 4.1 106

    tonne

    17.15 Zero.

    17.17 (a) 2 Cr(s) + 3 Cl2(g) 2 CrCl3(s)

    (b) Cr(s) + 2 ICl(l) CrCl2(s) + I2(s)

    17.19 Iron(III) iodide will not be stable because iodide ionis a reducing agent.

    17.21 H = 7838 kJ. It would be a good propellantbecause it produces a large number of small gas

    molecules.

    17.23 10 H2S(g) + 6 I2O5(s) 10 SO2(g) + 6 I2(s) + 10

    H2O(l)I2(s) + 2 S2O3

    2(aq) 2 I(aq) + S4O62(aq)

    17.25 Steric hindrance.

    17.27

    17.29 It would start to show some metallic properties; thediatomic element might be a significant electrical

    conductor; common oxidation state of 1; form an

    insoluble compound with silver ion. Astatine shouldform interhalogen compounds.

    17.31 Structure (c), with the charge on the sulfur atom,

    must be the major contributor.

    17.33 Fluorine:

    Cl2(g) + 3 F2(g) 2 ClF3(g)

    S(s) + 3 F2(g) SF6(g)

    BrO3(aq) + F2(g) + 2 OH

    (aq) BrO4(aq) + 2

    F(aq) + H2O(l)

    2 Fe(s) + 3 F2(g) 2 FeF3(s)

    H2(g) + F2(g) 2 HF(g)

    2 F(KH2F3) F2(g) + 2 e

    HF(aq) + OH(aq) H2O(l) + F(aq)

    HF(aq) + F(aq) HF2

    (aq)

    6 HF(aq) + SiO2(s) SiF62(aq) + 2 H+(aq) + 2

    H2O(l)

    4 HF(g) + UO2(s) UF4(s) + 2 H2O(g)

    UF4(s) + F2(g) UF6(g)Chlorine:

    P4(s) + 10 Cl2(g) 4 PCl5(s)

    2 Fe(s) + 3 Cl2(g) 2 FeCl3(s)

    3 Cl2(g) + NH3(g) NCl3(l) + 3 HCl(g)

    Cl2(aq) + 2 OH(aq) Cl(aq) + ClO(aq) + H2O(l)

    ClO(aq) + H+(aq) HClO(aq)

    2 ClO(aq) + Ca2+(aq) Ca(ClO)2(s)

    Cl2(g) + H2(g) 2 HCl(g)

    2 HCl(g) + Fe(s) FeCl2(s) + H2(g)

    3 Cl2(aq) + 6 OH(aq) ClO3

    (aq) + 5 Cl

    (aq) + 3

    H2O(l)

    ClO3(aq) + H2O(l) ClO4

    (aq) + 2 H+(aq) + 2 e

    2 ClO3(aq) + 4 H+(aq) + 2 Cl(aq) 2 ClO2(aq) +Cl2(g) + 2 H2O(l)

    Iodine:

    I2(s) + Cl2(g) 2 ICl(s)

    I2(s) + 2 S2O32(aq) 2 I(aq) + S4O6

    2(aq)

    2 I(aq) + Cl2(g) I2(aq) + 2 Cl(aq)

    I(aq) + I2(aq) I3

    (aq)

    17.35 Chlorine oxidation state = +1, oxygen = 1.

    17.37 The iodide anion will stabilize the large low-charge

    cation.

    17.39 BrF would be an analog of Cl2.

    17.41 (a) (CN)2; (b) AgCN, or Pb(CN)2, or Hg2(CN)2.

    17.43 P(CN)3

    Beyond the Basics

    17.45 The ammonium hydrogen fluoride may bedecomposing.

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    17.47

    17.49 Dichlorine heptaoxide. It is the oxide in the higher

    oxidation state.

    17.51 The bond angles will be approximately 109.

    17.53 2 NH4ClO4(s) N2(g) + Cl2(g) + 2 O2(g) + 4 H2O(g)

    17.55

    17.57 Tl+(I3). Iodide is a reducing agent.

    17.59 (a) The azide (N3) ion, acts as a pseudohalide ion.

    Thus it can form a pseudo-interhalide ion.(b) Higher.

    (c) There will be a trigonal bipyramid electron-pair

    arrangement.

    (d) By a large cation.

    17.61 (a) ClF3(l) + BF3(g) ClF2+(ClF3) + BF4

    (ClF3)

    (b) ClF3(l) + KF(s) K+(ClF3) + ClF4

    (ClF3)

    (c) In (a), the BF bond is much stronger than the Cl

    F bond. In (b), the ClF bond strength must begreater than the energy needed to extract a

    fluoride ion from the potassium fluoride lattice.

    Chapter 18

    Exercises

    18.1 (a) Xe(g) + 2 F2(g) XeF4(s)

    (b) XeF4(s) + 2 PF3(g) 2 PF5(g) + Xe(g)

    18.3 Descending, the melting and boiling points increase,

    as do the densities.

    18.5 Helium cannot be solidified under normal pressure;

    when cooled close to absolute zero, liquid helium

    becomes an incredible thermal conductor.18.7 The bond order must be .

    18.9 The weakness of the fluorine-fluorine bond that has

    to be broken, and the comparative strength of the

    xenon-fluorine bond.

    18.11

    18.13 The double-bonded structure probably makes a majorcontribution to the bonding.

    18.15 Using the calculation method:

    (a) +4

    (b) +6(c) +8

    18.17 Rubidium or cesium.

    18.19 2 Au + 7 KrF22 (KrF)+(AuF6

    ) + 5 Kr

    18.21 Xe(g) + F2(g) XeF2(s)

    2 XeF2(s) + 2 H2O(l) 2 Xe(g) + O2(g) + 4 HF(l)

    Xe(g) + 2 F2(g) XeF4(s)

    Xe(g) + 3 F2(g) XeF6(s)

    XeF6(s) + H2O(l) XeOF4(l) + 2 HF(l)

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    XeOF4(l) + 2 H2O(l) XeO3(s) + 4 HF(l)

    XeO3(s) + OH(aq) HXeO4

    (aq)

    2 HXeO4(aq) + 2 OH

    (aq) XeO6

    4(aq) + Xe(g) +O2(g) + H2O(l)

    XeO64(aq) + 2 Ba2+(aq) Ba2XeO6(s)

    Ba2XeO6(s) + 2 H2SO4(aq) 2 BaSO4(s) + XeO4(g)

    + 2 H2O(l)

    Beyond the Basics

    18.23 XeF2(SbF5) + SbF5(l) XeF+(SbF5) + SbF6

    (SbF5)

    18.25 The ArF bond energy = 77.5 kJmol-1.

    Chapter 19

    Exercises

    19.1 (a) Element belonging to the d-block.

    (b) Molecules or ions covalently bonded to a centralmetal ion.

    (c) Energy separation between different members ofthe metals d-orbital set.

    19.3 The cyanide ligand stabilizes low oxidation states and

    stabilizes normal ones.

    19.5 [Pt(NH3)4]2+[PtCl4]

    2

    19.7 The geometric isomers are:

    There are two optical (chiral) isomers.

    19.9 (a) Ammonium pentachlorocuprate(II); (b)pentaammineaquacobalt(III) bromide; (c) potassium

    tetracarbonylchromate(-III); (d) potassium

    hexafluoronickelate(IV); (e) tetraamminecopper(II)

    perchlorate.

    19.11 (a) [Mn(OH2)6](NO3)2, (b) Pd[PdF6], (c)

    [CrCl2(OH2)4]Cl2 H2O, (d) K3[Mo(CN)8].

    19.13 (a) The d6configuration in an octahedral field:

    (b) The d6configuration in a tetrahedral field:

    19.15 The largest value of is for the cobalt(III) complex,the others being cobalt(II) because the splitting

    increases with increase in oxidation state.

    19.17 (a) [ReF6]2, the heavier metal has greater

    crystal field splitting.(b) [Fe(CN)6]

    3, the higher charge has greater

    crystal field splitting.

    19.19 ConfigurationCFSE: d0, 0.0 tet, ascending to d2,

    1.2 tet, descending to d5, 0.0 tet, repeating to

    d10, 0.0 tet.

    19.21 Normal spinel, because the Cr3+ion will have a

    greater CFSE than that of the Ni2+ion.

    19.23 [Ni(OH2)6]2+(aq) + 2 det(aq) [Ni(det)2]

    2+(aq) +

    6 H2O(l)

    The chelate effect.

    Beyond the Basics

    19.25 The ligand is probably too large to fit in addition to

    the three chloro-ligands.

    19.27 (a) M2+should disproportionate as the sum of the

    potentials is positive. 3 M2+(aq) M(s) + 2

    M3+(aq)

    (b) pH = 3.38

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    19.29 For zinc, with its filled d10orbitals, there is no CFSE.

    For nickel, a square-planar geometry will maximize

    CFSE and it will enable some degree of bonding tooccur.

    19.31 (a) [Cr(OH2)6]3+

    3Cl, hexaaquachromium(III)

    chloride;(b) [Cr(OH2)5Cl]2+2Cl

    ,

    pentaaquachlorochromium(III) chloride;

    (c) [Cr(OH2)4Cl2]+Cl,

    tetraaquadichlorochromium(III) chloride.

    19.33 Fluoride is a weaker field ligand than chloride.

    Chapter 20

    Exercises

    20.1 (a) TiCl4(l) + O2(g) TiO2(s) + 2 Cl2(g)

    (b) Na2Cr2O7(s) + S(l) Cr2O3(s) + Na2SO4(s)

    (c) Cu(OH)2(s) CuO(s) + H2O(l)

    20.3 For the earlier part of the Period 4 elements, themaximum oxidation number is the same as the group

    number. For the later members, the oxidation state of

    +2 predominates.

    20.5 Titanium(IV) chloride vaporizes readily.

    20.7 (a) MnO4(aq) + 8 H+(aq) + 5 e

    Mn2+(aq) + 4 H2O(l)

    (b) MnO4(aq) + 2 H2O(l) + 3 e

    MnO2(s)

    + 4 OH(aq)

    20.9 Fe(s) + 2 HCl(g) FeCl2(s) + H2(g)

    2 Fe(s) + 3 Cl2(g) 2 FeCl3(s)

    20.11 (a) Cobalt, (b) Copper, (c) Chromium.

    20.13 The two reactants are the hexaaquairon(III) ion and

    thiosulfate ion:

    Fe3+(aq) + 2 S2O32(aq)

    [Fe(S2O3)2](aq)

    [Fe(S2O3)2](aq) + Fe3+(aq) 2

    Fe2+(aq) + S4O62(aq)

    20.15 (a) Fluoride stabilizes high oxidation states.(b) Low spin.

    20.17 2 FeO42(aq) + 2 NH3(aq) + 10 H

    +(aq) 2 Fe3+(aq)+ N2(g) + 8 H2O(l)

    20.19 Chromium(VI) oxide. The very high charge density

    of the chromium metal ion will result in covalent

    bond formation.

    20.21 Chromium(III) ion will lose a hydrogen ion to awater molecule.

    20.23 According to Fajans Rules, cations with non-noble-gas configurations are likely to have a more covalent

    character.

    20.25 (a) FeO(OH), (b) Fe3+, (c) Fe2+.

    20.27 They both form anhydrous chlorides that react with

    water. In the gas phase, their chlorides exist as

    dimers, Al2Cl6and Fe2Cl6. On the other hand,iron(III) oxide is basic, while the oxide of aluminum

    is amphoteric.

    20.29 Titanium:

    TiO2(s) + 2 C(s) + 2 Cl2(g) TiCl4(g) + 2

    CO(g)

    TiCl4(g) + O2(g) TiO2(s) + 2 Cl2(g)

    TiCl4(g) + 2 Mg(l) Ti(s) + 2 MgCl2(l)

    Vanadium:

    [H2VO4](aq) + 4 H+(aq) + eVO2+(aq) + 3

    H2O(l)

    VO2+(aq) + 2 H+(aq) + eV3+(aq) + H2O(l)

    [V(OH2)6]3+(aq) + e[V(OH2)6]

    2+(aq)

    Chromium:

    CrO42(aq) + 2 Ag+(aq) Ag2CrO4(s)

    CrO42(aq) + H2O(l) HCrO4

    (aq) + OH(aq)

    2 CrO42(aq) + 2 H+(aq) Cr2O7

    2(aq) + H2O(l)

    Cr2O72(aq) + 2 NH4

    +(aq) (NH4)2Cr2O7(s)

    (NH4)2Cr2O7(s) Cr2O3(s) + N2(g) + 4 H2O(l)

    Cr2O72(aq) + 14 H+(aq) + 6 e2 Cr3+(aq) + 7

    H2O(l)

    Cr2O72(aq) + 2 K+(aq) K2Cr2O7(s)

    K2Cr2O7(s) + H2SO4(aq) 2 CrO3(s) + K2SO4(aq) +

    H2O(l)

    K2Cr2O7(s) + 4 NaCl(s) + 6 H2SO4(l) 2 CrO2Cl2(l)+ 2 KHSO4(s)

    + NaHSO4(s) + 3 H2O(l)

    CrO2Cl2(l) + 4 OH

    (aq) CrO42

    (aq) + 2 Cl

    (aq) +2 H2O(l)

    Cr2O72(aq) + 14 H+(aq) + 6 e2 Cr3+(aq) + 7

    H2O(l)

    2 Cr3+(aq) + Zn(s) 2 Cr2+(aq) + Zn2+(aq)

    2 Cr2+(aq) + 4 CH3COO-(aq) + 2 H2O(l)

    Cr2(CH3COO)4(OH2)2(s)

    Manganese:

    MnO4(aq) + eMnO4

    2(aq)

    MnO42

    (aq) + 2 H2O(l) + 2 eMnO2(s) + 4

    OH(aq)

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    MnO4(aq) + 2 H2O(l) + 3 e

    MnO2(s) + 4

    OH(aq)

    MnO4(aq) + 8 H+(aq) + 5 e

    Mn2+(aq) + 4 H2O(l)

    Mn2+(aq) + 2 OH(aq) Mn(OH)2(s)

    Mn(OH)2(s) + OH(aq) MnO(OH)(s) + H2O(l) +

    e

    Iron:[Fe(OH2)6]

    3+(aq) + SCN(aq) [Fe(SCN)(OH2)5]

    2+(aq) + H2O(l)

    [Fe(OH2)6]3+(aq) + 4 Cl(aq) [FeCl4]

    (aq) + 6H2O(l)

    Fe3+(aq) + 3 OH(aq) FeO(OH)(s) + H2O(l)

    [Fe(OH2)6]3+(aq) + e[Fe(OH2)6]

    2+(aq)

    Fe3+(aq) + 2 S2O32(aq) [Fe(S2O3)2]

    (aq)

    [Fe(S2O3)2](aq) + Fe

    3+(aq) 2 Fe

    2+(aq) +

    S4O62(aq)

    Fe2+(aq) + 2 OH(aq) Fe(OH)2(s)

    [Fe(OH2)6]2+(aq) + NO(aq) [Fe(NO)(OH2)5]

    2+(aq)+ H2O(l)

    Fe(OH)2(s) + OH(aq) FeO(OH)(s) + H2O(l) + eFe2+(aq) + 2 eFe(s)

    2 Fe(s) + 3 Cl2(g) 2 FeCl3(s)

    Fe(s) + 2 HCl(g) FeCl2(s) + H2(g)

    Cobalt:

    [Co(OH2)6]3+(aq) + e[Co(OH2)6]

    2+(aq)

    [Co(OH2)6]2+

    (aq) + 4 Cl(aq) [CoCl4]

    2(aq) + 6

    H2O(l)

    Co2+(aq) + 2 OH(aq) Co(OH)2(s)

    Co(OH)2(s) + 2 OH(aq) Co(OH)4

    2(aq)

    Co(OH)2(s) + OH(aq) CoO(OH)(s) + H2O(l) + e

    [Co(OH2)6]2+(aq) + 6 NH3(aq) [Co(NH3)6]

    2+(aq) +

    6 H2O(l)[Co(NH3)6]2+(aq) [Co(NH3)6]

    3+(aq) + e

    O2(g) + 2 H2O(l) + 4 e4 OH

    (aq)

    Nickel:

    Ni(CO)4(g) Ni(s) + 4 CO(g)

    Ni(s) Ni2+(aq) + 2 e

    [Ni(OH2)6]2+(aq) + 4 Cl(aq) [NiCl4]

    2(aq) + 6

    H2O(l)

    Ni2+(aq) + 2 OH(aq) Ni(OH)2(s)

    [Ni(OH2)6]2+(aq) + 6 NH3(aq) [Ni(NH3)6]

    2+(aq) +6 H2O(l)

    Copper:

    2 Cu(s) + 2 H+(aq) + 4 Cl(aq) 2 [CuCl2](aq) +

    H2(g)Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)

    Cu(s) Cu2+(aq) + 2 e

    [Cu(OH2)6]2+(aq) + 4 NH3(aq) [Cu(NH3)4]

    2+(aq) +6 H2O(l)

    [Cu(OH2)6]2+(aq) + 4 Cl(aq) [CuCl4]

    2(aq) + 6H2O(l)

    Cu2+(aq) + 2 OH(aq) Cu(OH)2(s)

    Cu(OH)2(s) + 2 OH(aq) [Cu(OH)4]

    2(aq)

    Cu(OH)2(s) CuO(s) + H2O(l)

    Beyond the Basics

    20.31 Addition of an anhydrous calcium compound will

    result in formation of the hexaaquacalcium ion.

    Addition of an anhydrous zinc compound results in

    the formation of the competing complexation.

    20.33 Cr2O72

    (aq) + H2O(l) 2 CrO42

    (aq) + 2 H

    +

    (aq)Pb2+(aq) + CrO42(aq) PbCrO4(s)

    20.35 (a) Nickel(II) hydroxide.

    (b) This should be the square planar

    tetracyanonickelate(II) ion.

    (c) This must involve the addition of a fifth cyanideion.

    20.37 (a) The high-charge cation (Fe3+) will have asomewhat low lattice energy when

    combined with a low-charge anion (ClO4).

    (b) Either: Ammonia and water are quite high inthe spectrochemical series. Or: Ammonia

    and water are hard bases.

    (c) Bromide is more easily reduced than

    chloride; thus the charge transfer takes place

    at a lower energy.

    20.39 Ni = +2, S = 1.

    20.41 [A] Nickel(II) sulfide; [B] hydrogen sulfide; [C]hexaaquanickel(II) ion;

    [D] sulfur dioxide; [E] sulfur; [F] and [G] disulfur

    dichloride and sulfur dichloride; [H]hexaamminenickel(II) ion; [I] nickel(II) hydroxide;

    [J] nickel metal; [K] tetracarbonylnickel(0).

    NiS(s) + 2 H+(aq) Ni2+(aq) + H2S(g)

    2 H2S(g) + 3 O2(g) 2 H2O(l) + 2 SO2(g)2 H2S(g) + SO2(g) 2 H2O(l) + 3 S(s)

    2 S(s) + Cl2(g) S2Cl2(l)

    S(s) + Cl2(g) SCl2(l)

    [Ni(OH2)6]2+(aq) + 6 NH3(aq) [Ni(NH3)6]

    2+(aq) +

    6 H2O(l)

    Ni2+(aq) + 2 OH(aq) Ni(OH)2(s)

    Ni2+(aq) + Zn(s) Ni(s) + Zn2+(aq)

    Ni(s) + 4 CO(g) Ni(CO)4(l)

    20.43 Vanadium.

    20.45 This corresponds to a full neutron shell.

    20.47 3+, as the shared oxygen would have an oxidation

    state of 2. The linear shape suggests there is a -

    bonding CrOCr system.

    20.49 Presumably the chloride ligand has preferentially

    stabilized the 3+ oxidation state of the iron.

    20.51 As the halide ion is more readily oxidized, theabsorption of light will be more and more in the

    visible part of the spectrum.

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    20.53 Calcium will replace the Mn2+. Iron would most

    likely replace the Mn3+. Titanium would replace the

    silicon. Aluminum could replace the Mn3+.

    20.55 Some form of -bonding through the dorbitals.

    20.57 (a) Fe(s) + O2(g) Fe2O3(s)(b) Sodium silicate prevents the continuation of theoxidation.

    (c) The red-hot iron would have reacted with water to

    give hydrogen gas. The explosion would have

    resulted from a hydrogen/oxygen mixture.

    Chapter 21

    Exercises

    21.1 (a) 2 [Ag(CN)2](aq) + Zn(s) 2 Ag(s) +

    [Zn(CN)4]2(aq)

    (b) 2 Au(s) + 3 Cl2(g) 2 AuCl3(s)

    21.3 Discussing the 5dfluorides, the oxidation number

    seems to plateau at seven.

    21.5 (a) automobile engine lubricant; (b) antibacterial.

    21.7 Osmium(VIII) oxide has a melting point of 40C, and

    is very soluble in low-polarity, organic solvents.

    21.9 Ruthenium, rhodium, palladium, osmium, iridium,and platinum.

    21.11 The 3dtransition metals tend to have lower oxidationstates than those of the 4dand 5dseries. The smaller

    3dions cannot accommodate as many ligands.

    21.13 For Pd: +2 and +3. For Pt: +2, +4, and +6. Square

    planar is common for the lower oxidation states,

    octahedral geometry for the +6.

    21.15 End-on overlap of a pair of orbitals; diagonal

    overlap of a pair of orbitals; and the side-to-side

    overlap of a pair of orbitals.

    21.17 PdF3has the formulation of: (Pd2+)[PdF6]

    2.

    21.18 It has a stable, water-soluble, species at near-neutralpH making it transportable by biological fluids.

    Beyond the Basics

    21.20 Fluorine tends to promote metals to their highest

    oxidation states. WF6.

    21.22 The potassium halides are all water-soluble while all

    of the silver halides are insoluble. Low-charge-

    density cations result in low lattice energies and such

    salts should be water soluble.

    21.24 Though thorium is an actinoid, the early actinoids

    favor oxidation states matching their analogous groupnumber.

    21.26 In the complex shown, each iodide bridges threeniobium atoms. [Nb6I8]

    3+

    21.28 The Re3Cl9structure involves a central triangle of

    rhenium atoms with bridging and terminal chlorineatoms in a polymeric structure.

    Chapter 22

    Exercises

    22.1 (a) Zn(s) + Br2(l) ZnBr2(s)

    (b) ZnCO3(s) ZnO(s) + CO2(g)

    22.3 Zn(s) + 2 H+(aq) Zn2+(aq) + H2(g)

    Zn2+(aq) + CO32(aq) ZnCO3(s)

    22.5 (a) Zinc and magnesium have the following

    similarities: their cations are 2+ ions of similar size,they are colorless, and they both form hexahydrates.

    Both elements form soluble chlorides and sulfates,

    and insoluble carbonates.

    (b) The only two common features are that both zinc

    and aluminum are amphoteric metals, reacting withboth acids and bases, and they are both strong Lewis

    acids.

    22.7 Cd(OH)2(s) + 2 eCd(s) + 2 OH(aq)

    2 Ni(OH)2(s) + 2 OH(aq) 2 NiO(OH)(s) + 2

    H2O(l) + 2 e

    22.9 Cadmium metal was used as a coating for paper clips

    primarily because it was a sacrificial anode. As

    cadmium compounds are highly toxic, cadmiumplating has been discontinued.

    22.11 Zn(s) + 2 H+(aq) Zn2+(aq) + H2(g)

    Zn(OH2)62+(aq) + 4 NH3(aq) Zn(NH3)42+(aq) + 6H2O(l)

    Zn2+

    (aq) + 2 OH(aq) Zn(OH)2(s)

    Zn(OH)2(s) + 2 OH(aq) Zn(OH)4

    2(aq)

    Zn(OH)2(s) ZnO(s) + H2O(l)

    ZnO(s) + 2 H+(aq) Zn2+(aq) + H2O(l)

    ZnCO3(s) ZnO(s) + CO2(g)

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    Beyond the Basics

    22.13 Mercury(I) undergoes a disproportionation

    equilibrium.

    22.15 The metals are very different in size.

    22.17 Sulfur. Mercury(II) is a soft acid. Sulfur is a soft

    base.

    22.19 (a) Zn(NH2)2(NH3) + 2 NH4+(NH3)

    Zn(NH3)42+(NH3)

    (b) Zn(NH2)2(NH3) + 2 NH2(NH3) Zn(NH2)4

    2

    (NH3)

    22.21 Zinc oxide.

    22.23 Hydrogen sulfide is in a two-step equilibrium withthe sulfide ion. When acidified, the increased

    hydronium-ion concentration will drive the

    equilibria to the left.

    Chapter 23

    Exercises

    23.1 (a) organometallic

    (b) not organometallic as the bond B-O not B-C(c) organometallic

    (d) not organometallic as nitrogen is not metallic

    (e) not organometallic as there is no Na-C bond(f) organometallic

    (g) organometallic

    23.3 (a) Bi(CH3)5

    (b) Si(C6H5)4 tetraphenyl silane

    (c) KB(C6H5)4postassium teraphenylborane(d) Li4(CH3)4(e) (C2H5)MgCl

    23.5 C2H5MgBr will be tetrahedral with two molecules of

    solvent coordinated to the magnesium.

    23.7 Hg(CH3)2+ 2 Na 2 NaCH3+ Hg

    23.9 (a) LiCH3+ LiBr

    (b) 2 LiCl + Mg(C2H5)2(c) Mg(C2H5)2+ Hg(d) Li(C6H5) + C2H6

    (e) C2H5MgCl + Hg(f) B(CH2CH2CH3)3

    (g) Sn(C2H5)4+ 4 MgCl2

    23.11 (a) hexacarbonylchromium(0)

    (b)ferrocene orbis(pentahaptocyclopentadienyl)iron(II)

    (c)hexahaptobenzenetricarbonylchromium(0)

    (d)pentahaptocyclopentadienyltricarbonyltungsten(I)

    (e)bromopentacarbonylmanganese(I)

    23.13 Cr(CO)6

    Fe(CO)5

    Ni(CO)4

    Mn

    COCO

    COOC

    OC Mn

    COCO

    COOC

    CO

    23.15 V(CO)6is a seventeen electron complex.

    23.17 (a) 1 MnMn bond

    (b) 2 MnMn bonds

    Mn Mn

    CO

    OCCO

    OC

    (c) 1 FeFe bond

    Fe Fe

    COCOOCCO

    CO

    (d) no Mn-Mn bonds

    Mn Mn

    Br

    Br

    CO

    CO

    COCO

    OC

    OC

    OCOC

    23.19 (a) [Cr(CO)6] + 3 CH3CN [Cr(CO)3(CH3CN)3] + 3 CO

    (b) [Mn2(CO)10] + H22 [HMn(CO)5]

    (c) [Mo(CO)6] +

    (CH3)2PCH2CH2P(Ph)CH2CH2P(CH3)2[Mo(CO)3((CH3)2PCH2CH2P(Ph)CH2CH2P(CH3)2)] +

    3 CO

    (d) [Fe(CO)5] + 1,3-cyclohexadiene 2 CO +

    (CO)3Fe

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    Descriptive Inorganic Chemistry, Fifth Edition Answers to Odd-Numbered Questions 25

    (e) 23.29

    (f)

    (g)

    [PtCl2(PMe3)2] + LiCH2CH2CH2CH2Li 2 LiCl+ (PMe3)2Pt

    (h) [Ni(CO)4]+ PF3 [Ni(CO)3PF3] + CO

    (i)

    [Mn2(CO)10] + Br22 [Mn(CO)5Br]

    (j) [HMn(CO)5] + CO2[(CO)5MnCOOH]

    23.21 (a) +3

    (b)+1

    Beyond the Basics

    23.23

    (5-C5H5)2Ni + Ni(CO)4 Ni Ni

    OC

    CO

    2

    23.25

    A = tricarbonyl(5-cyclopentadienyl)(1-propenyl)tungsten(II)

    B = dicarbonyl(5-cyclopentadienyl)(3-propenyl)tungsten(II)

    C = tricarbonyl(5-cyclopentadienyl)(2-propenyl)tungsten(II)

    hexafluorophosphate

    Evolved gas = propene

    23.27 Ti(S2CEt2)4.